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Thread: Finding Ker(f)

  1. #1
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    Finding Ker(f)

    Hello !
    I'm a bit confused about the notion of Ker() and the solution I was given for an exercise.

    f is a linear application defined for u = (x,y,z) by :

    f(u) = (x + y + z, x + 2y + 3z, x + 3y + 5z)

    Written as a system of equations :

     \left\{\begin{matrix} x + y + z = 0<br />
\\ x + 2y + 3z = 0<br />
\\ x + 3y + 5z = 0<br /> <br />
\end{matrix}\right.<br />

    I was given as a solution Ker(f) = Vect{v1} = 1 with v1 =(-2,-3,1)

    However, I have no idea how to get to this vector.
    I solved the system by putting it in a rref matrix and got as solution x = z, y = -2z.
    So this would give me v1 = (1,-2,1) if z equals one ? So I get Ker(f) = 1 also but I don't get why my vector is different and not even collinear to the one from the solution.

    I also have trouble understanding how to interpret what Ker() or Im() represent (geometrically or just understand the concept).

    Thank you for your help !
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  2. #2
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    Re: Finding Ker(f)

    The image of a function is the set of all possible values the function can take, which is a subset of the function's codomain. The kernel of a function is the set of all possible inputs to a function that the function will take to zero, so the kernel is a subset of the domain of the function.

    Your vector is correct. Try plugging in the vector you were given as a solution, and you will find that it does not yield the zero vector.
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  3. #3
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    Re: Finding Ker(f)

    Quote Originally Posted by Jo37 View Post
    Hello !
    I'm a bit confused about the notion of Ker() and the solution I was given for an exercise.

    f is a linear application defined for u = (x,y,z) by :

    f(u) = (x + y + z, x + 2y + 3z, x + 3y + 5z)

    Written as a system of equations :

     \left\{\begin{matrix} x + y + z = 0\\ x + 2y + 3z = 0\\ x + 3y + 5z = 0\end{matrix}\right.

    I was given as a solution Ker(f) = Vect{v1} = 1 with v1 =(-2,-3,1)

    However, I have no idea how to get to this vector.
    I solved the system by putting it in a rref matrix and got as solution x = z, y = -2z.
    So this would give me v1 = (1,-2,1) if z equals one ? So I get Ker(f) = 1 also but I don't get why my vector is different and not even collinear to the one from the solution.
    I also have trouble understanding how to interpret what Ker() or Im() represent (geometrically or just understand the concept).
    Look at this.
    Null space is another name for kernel.
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  4. #4
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    Re: Finding Ker(f)

    Quote Originally Posted by Jo37 View Post
    I was given as a solution Ker(f) = Vect{v1} = 1 with v1 =(-2,-3,1)
    I have no idea what this means! If v1 is a vector what does "Vect{v1}" mean? The vector space spanned by v1? But then what does "Vect{v1}= 1" mean? How is a vector space equal to 1? In any case, "(-2, 3, 1)" is not in the kernel. In particular, x+ y+ z= -2+ 3+ 1= 2, not 0!

    Given x + y + z = 0, x + 2y + 3z = 0, x + 3y + 5z = 0, we can subtract the first equation from the second to get y+ 2z= 0 and subtract the first equation from the third to get 2y+ 4z= 0. From the first of those, y= -2z so 2y= -4z and the last equation becomes -4z= -4z, true for all z. With y= -2z, x+ y+ z= x- 2z+ z= x- z= 0 so x= z. That is, (x, y, z)= (z, -2z, z) satisfies all three equations for any number, z. We can write that as (z, -2z, z)= z(1, -2, 1). The kernel is the one dimensional subspace of all multiples of (1, -2, 1).
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  5. #5
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    Re: Finding Ker(f)

    Thank you very much for your answers !

    @HallsofIvy This is literally what was written in the solution I was given. How to write the solution properly ? Ker(f) = (z, -2z, z) and dim(Ker) = 1 ?
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  6. #6
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    Re: Finding Ker(f)

    Quote Originally Posted by Jo37 View Post
    Thank you very much for your answers !

    @HallsofIvy This is literally what was written in the solution I was given. How to write the solution properly ? Ker(f) = (z, -2z, z) and dim(Ker) = 1 ?
    I would prefer to write "Ker(f)= {(z, -2z, z)}"- the set of all such vectors. Or you could write that the kernel is the space of all multiples of (1, -2, 1).
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